IEER SDA Vol. 5 No. 4

Answers to:
The Atomic Puzzler
SDA Vol. 5 No. 3


1. How much plutonium would be in the spent fuel if none of the new plutonium made from U-238 were fissioned? (Round to the nearest whole percentage point.)

The MOX fuel contains 4% plutonium and 96% uranium. Since the reactor has a 100% MOX core, then the reactor also contains 4% plutonium and 96% uranium. Since three-forths of the original plutonium is converted to fission products, then one-forth of the plutonium remains: So:

0.25 x 0.04 (Pu) = 0.01 (the spent fuel contains 1% Pu)

In addition, 2.1% (0.021) of the uranium is also converted to plutonium, so:

0.021 x 0.96 (U-238) = 0.02 (2% additional Pu in spent fuel - rounded from 0.02016)


Add these to get the total amount of plutonium left in the spent fuel:

0.01 (Pu) + 0.02 (Pu from U-238) = 0.03 (three percent)

Therefore, the spent fuel would contain three percent (3%) plutonium.


2. How much plutonium would be left in the spent fuel if half of the new plutonium made from U-238 is fissioned also? (Round to the nearest whole percentage point.)

From the first question, we know there is 1% plutonium in the spent fuel remaining from the original plutonium. If half of the plutonium from U-238 is fissioned also, then:

0.021 (Pu from U-238) x 0.5 = 0.01
(1% additional Pu in spent fuel - rounded from 0.0105 )

Repeating step 3 above we get:

0.01 (Pu) + 0.01 (Pu from U-238) = 0.02 (two percent)

Therefore, the spent fuel would contain two percent (2%) plutonium.


3. Gamma estimated that in Case 1 above, the spent fuel would have 3% fission products. Is he right? What percent of fission products exists in the spent fuel in Case 2? (For Case 2 round to the nearest whole percentage point.)

In Case 1, we knew that three-forths of the plutonium was converted to fission products. So:

0.75 x 0.04 = 0.03 (three percent) So, Gamma is correct!

In Case 2, we still have 3% fission products from the plutonium. We simply need to add the fission products from the fissioning of the plutonium from U-238:

0.02 ( Pu from U-238) x 0.5 = 0.01 (one percent fission products)

So:

0.03 (fission products from Pu)
+ 0.01 (fission products from Pu from U-238)
____
= 0.04 (four percent)

Therefore, the spent fuel would contain four percent (4%) fission products.


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Institute for Energy and Environmental Research

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May, 1997